NEET2010PhysicsMagnetic Properties of MatterActual
A vibration magnetometer placed in magnetic meridian has a small bar magnet. The magnet executes oscillations with a time period of 2 ~s in earth's horizontal magnetic field of 24 T . When a horizontal field of 18 T is produced opposite to the earth's field by placing a current carrying wire, the new time period of magnet will be
Options
- A1 ~s
- B2 ~s
- C3 ~s
- D4 ~s
Correct answer
D. 4 ~s
Step-by-step solution
The time period T of oscillation of a magnet is given by (T=2 I M B ) where, I = Moment of inertia of the magnet about the axis of rotation (M= ) Magnetic moment of the magnet (B= ) Uniform magnetic field As I, B remains the same ( T 1 B ) or ( T₂ T₁ = B₁ B₂ ) According to given problem, ( aligned B₁ & =24 T B₂ & =24 T -18 T =6 T T₁ & =2 ~s T₂ & =(2 ~s ) (24 T ) (6 T ) =4 ~s aligned )